Let S_1=\{x \in R-\{1,2\}: \frac{(x+2)(x^2+3 x+5)}{-2+3 x-x^2} \geq 0\} and S_2=\{x \in R: 3^{2 x}-3^{x+1}-3^{x+2}+27 \leq 0\}. Then, S_1 \cup S_2 is equal to : [30-Jun-2022-Shift-1]

Correct Answer is : (−∞,−2]∪[1,2]

Quadratic Equation and Inequalities Part 2

Section: Mathematics

Question : 78 of 93

Marks: +1, -0

Let S1={x∈R−{1,2}:‌

(x+2)(x2+3x+5)

−2+3x−x2

≥0} and S2={x∈R:32x−3x+1−3x+2+27≤0}. Then, S1∪S2 is equal to :

[30-Jun-2022-Shift-1]

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